Problem 19
Question
You are asked to prepare a buffer solution with a pH of 3.50. The following solutions, all \(0.100 \mathrm{M},\) are available to you: HCOOH, CH \(_{3} \mathrm{COOH}, \mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{NaCHOO}\) \(\mathrm{NaCH}_{3} \mathrm{COO},\) and \(\mathrm{NaH}_{2} \mathrm{PO}_{4} . \quad\) Describe how you would prepare this buffer solution. [Hint: What volumes of which solutions would you use?
Step-by-Step Solution
Verified Answer
To make a buffer solution of pH 3.50, we should mix the formic acid HCOOH and sodium formate NaHCOO solutions, taking 1 liter of formic acid for every 0.58 liters of sodium formate solution.
1Step 1: Choosing the right solution to create a buffer
From the given \(0.100 \mathrm{M}\) solutions, the correct one to choose would be the one that yields a buffer with a pH of 3.50. We choose the weak acid and its conjugate base fron the list provided: HCOOH (Formic acid) and NaCHO (\( \text {Sodium formate}\))
2Step 2: Calculate the dissociation constant for Formic acid
The first step in preparing a buffer solution is to find the dissociation constant, pKa, of the chosen weak acid from a standard chemistry data tables. pKa of a substance is related to pH and indicates how readily hydrogen ions are given off by the acid in the solution. For Formic acid the pKa value is 3.74.
3Step 3: Applying the Henderson-Hasselbalch equation
For preparing a buffer solution, the Henderson-Hasselbalch equation is used which is \(pH = pKa + log \left( \frac {[Base]}{[Acid]} \right)\). Here, [Base] indicates the concentration of the base (Sodium formate) and [Acid] indicates the concentration of the acid (Formic acid). Substituting the given and found values in the equation we have, 3.50 = 3.74 + log \left( \frac {[Base]}{[Acid]} \right). Now we can solve for the ratio [Base]/[Acid].
4Step 4: Calculating concentrations
From the previous step, we can find the ratio [Base]/[Acid] = 0.58. Now to find the concentrations, one can keep the acid concentration at 0.100 M, and accordingly adjust the base concentration to achieve the required ratio. In this case, the base concentration will be 0.58 x 0.100 M = 0.058 M.
5Step 5: Finding the final volumes
Now, to prepare a buffer solution with the desired pH, you should mix the solutions in the right proportions. Since the sodium formate is 0.100 M and we need it to be 0.058 M, we take 0.58 L of sodium formate solution for each 1 L of formic acid solution. So preparing a buffer solution then depends on the total volume needed. If 1 L of buffer solution is needed, use 0.58 L of sodium formate and 1 L of formic acid. If less or more is needed, use the same ratios.
Key Concepts
Henderson-Hasselbalch EquationWeak AcidpH Calculation
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a powerful tool in chemistry, particularly when dealing with buffer solutions. This equation helps predict the pH of a buffer solution, which consists of a weak acid and its conjugate base. By using this formula, \[pH = pK_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right)\]we can determine the pH based on the acid's dissociation constant, \(pK_a\), and the concentrations of the base and acid.
It's crucial because it directly relates the acid-base concentration ratio with the solution's pH. In practical applications, such as in laboratories or industries, adjusting the ratio of base to acid enables precise tuning of the pH to the desired level. This is particularly useful when the pH needs to remain stable, even if small amounts of an acid or base are added to the solution.
What's notable about using the Henderson-Hasselbalch equation is that it simplifies calculations. By knowing a base and its conjugate acid that fits the desired pH range, one can quickly adjust amounts to achieve exact pH control. Let's explore more about weak acids involved in these calculations next.
It's crucial because it directly relates the acid-base concentration ratio with the solution's pH. In practical applications, such as in laboratories or industries, adjusting the ratio of base to acid enables precise tuning of the pH to the desired level. This is particularly useful when the pH needs to remain stable, even if small amounts of an acid or base are added to the solution.
What's notable about using the Henderson-Hasselbalch equation is that it simplifies calculations. By knowing a base and its conjugate acid that fits the desired pH range, one can quickly adjust amounts to achieve exact pH control. Let's explore more about weak acids involved in these calculations next.
Weak Acid
A weak acid is a type of acid that only partially dissociates in water. This means that when dissolved, not all the acid molecules donate their hydrogen ions to water, resulting in an equilibrium between the undissociated and dissociated species. This characteristic affects how the acid behaves in a solution.
Weak acids remain mostly undissociated compared to strong acids, which completely ionize. Examples of weak acids include acetic acid (vinegar) and formic acid, which are common in everyday products or biological systems.
The concept of weak acids is vital in the context of buffers or equilibrium chemistry. Since they do not release all their hydrogen ions at once, they contribute to the stability of a buffer solution. In a buffer system made with a weak acid and its conjugate base, if an acidic or basic component is added, the solution can absorb these changes, helping maintain a nearly constant pH level.
Understanding weak acids, therefore, provides insight into how buffer systems work and how they help control environments within a narrow pH range, crucial for many biological and chemical processes.
Weak acids remain mostly undissociated compared to strong acids, which completely ionize. Examples of weak acids include acetic acid (vinegar) and formic acid, which are common in everyday products or biological systems.
The concept of weak acids is vital in the context of buffers or equilibrium chemistry. Since they do not release all their hydrogen ions at once, they contribute to the stability of a buffer solution. In a buffer system made with a weak acid and its conjugate base, if an acidic or basic component is added, the solution can absorb these changes, helping maintain a nearly constant pH level.
Understanding weak acids, therefore, provides insight into how buffer systems work and how they help control environments within a narrow pH range, crucial for many biological and chemical processes.
pH Calculation
Calculating pH is a fundamental skill needed for understanding buffer solutions and acid-base chemistry. The pH, or 'potential of Hydrogen,' is a scale used to measure the acidity or basicity of an aqueous solution. It ranges from 0 to 14, with lower numbers indicating acidic solutions and higher numbers indicating basic solutions.
For any buffer solution, calculating its pH involves understanding how concentrations of the acid and its conjugate base work together. In buffer solutions, pH calculations often rely on the Henderson-Hasselbalch equation, which uses the relationship between the acid's \(pK_a\) and the molar concentrations of the acid and base.
Here's a simple guide to performing pH calculations in buffer solutions:
By understanding how to calculate pH, you gain the ability to manipulate and stabilize solutions across various applications, ensuring optimal conditions for reactions or processes that depend on a specific pH.
For any buffer solution, calculating its pH involves understanding how concentrations of the acid and its conjugate base work together. In buffer solutions, pH calculations often rely on the Henderson-Hasselbalch equation, which uses the relationship between the acid's \(pK_a\) and the molar concentrations of the acid and base.
Here's a simple guide to performing pH calculations in buffer solutions:
- Identify the weak acid involved and look up its \(pK_a\).
- Measure the initial concentrations of the acid and its conjugate base.
- Apply the Henderson-Hasselbalch equation to find the pH.
By understanding how to calculate pH, you gain the ability to manipulate and stabilize solutions across various applications, ensuring optimal conditions for reactions or processes that depend on a specific pH.
Other exercises in this chapter
Problem 13
What is the \(\mathrm{pH}\) of a solution obtained by adding \(1.15 \mathrm{mg}\) of aniline hydrochloride \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}
View solution Problem 14
What is the pH of a solution prepared by dissolving \(8.50 \mathrm{g}\) of aniline hydrochloride \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \math
View solution Problem 21
\(\begin{array}{lll}\text { Given } & 1.00 & \mathrm{L}\end{array}\) of a solution that is \(0.100 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\
View solution Problem 22
Given \(125 \mathrm{mL}\) of a solution that is \(0.0500 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{2}\) and \(0.0500 \mathrm{M} \mathrm{CH}_{3} \mathrm{NH}_{3}^{+
View solution